Click here to load reader

View

0Download

0

Embed Size (px)

535

6.1 Graphing Parabolas

We already know how to graph basic parabolas which open up or down

and which can be shifted around. Now, we will generalize what we

know in order to include parabolas which open sideways (left or right)

and we will also be able to graph wide and narrow parabolas. We will

utilize the process of completing the square in order to put our

quadratics into graphing form, so you may want to review section 2.8 as

well. We begin by comparing our basic parabola π¦ = π₯2 with the basic

sideways parabola π₯ = π¦2. Notice that π₯ and π¦ are switched in these two equations. (This is not a coincidence and it is a property of inverse

relations, which we study in more detail in section 6.5.)

π¦ = π₯2

π₯ = π¦2

π₯

π¦

π₯

π¦ Notice that the graph of π₯ = π¦2 could also be thought of as the two graphs we

get by solving this equation for π¦:

π₯ = π¦2 implies π¦ = Β±βπ₯, with

π¦ = βπ₯ representing the top half of the

graph and π¦ = ββπ₯

representing the bottom half.

536

In the same way that we were able to shift the graph of π¦ = π₯2 around,

we can also perform the same kinds of transformations on π₯ = π¦2.

Consider the graph of π¦ = (π₯ β 2)2 + 1, for instance. We know that one way of obtaining this graph is by first graphing our basic graph,

π¦ = π₯2, and then shifting it to the right by 2 and up 1 to obtain the following graph:

Similarly, we could consider the graph of π₯ = (π¦ β 2)2 + 1, but to shift this, we need to keep in mind which shift is attached to which variable.

For example, here the β2 belongs to π¦, and since it is on the same side

as π¦, we end up going in the opposite direction (because if you solve for

π¦, you will end up adding 2 to the other side). Hence the graph will shift

up 2. The value of 1 belongs to π₯ here, and we will end up shifting in the

same direction since the equation is already solved for π₯. So the graph will shift to the right by 1:

π₯

π¦

π₯

π¦

One way to remember this process

that matches what you already know

is to move the opposite direction for

what is in the parentheses and the

same direction for what is outside of

the parentheses.

537

We will not need to spend much time on this because there is a more

general way to graph parabolas that will include the parabolas that are

wider and narrower that these two basic parabolas. In order to graph

ALL parabolas, we need to allow for more than shifting and reflecting

because there is also βstretching or shrinkingβ involved for the ones that

have a coefficient other than 1 for the square term. We can take care of

this βstretching or shrinkingβ by plotting a couple of points (one of these

has to be the vertex) and using symmetry to finish our graph.

If we are going to use this method, we must have the parabola in

graphing form and we can put the parabola in graphing form by

completing the square.

For parabolas that open up or down, standard graphing form is as

follows: π = π(π β π)π + π

From this form, we can read off the vertex (π, π) which is going to be the highest or lowest point on the parabola β the place at which it βturns

aroundβ. (Note that to get the vertex, you take the opposite sign of what

is in the parentheses and the same sign of what is outside the

parentheses, just as when you are shifting. It is for the same reason.)

You can also read off the axis of symmetry π = π, which is the vertical line that divides the parabola into two equal halves. It is the line that acts

like a mirror on one side of the parabola to give us the other side. The

axis of symmetry is not part of the graph of the parabola, but sometimes

it is requested in the homework problems.

You can also tell whether the graph opens up or down by looking at the

sign on π. If π is positive, then the graph opens up. If π is negative, then the graph opens down. This really just helps you double check that you

graphed it correctly, as plotting the points should make it open the

correct way if there are no numerical errors.

Once you read off the vertex, you can choose another value to plug in

for π₯. (Since the domain for these types of parabolas is βall real

538

numbersβ, you can choose any value that you like). Plug your chosen

value into the equation to get the corresponding π¦ value and then plot

your point (π₯, π¦). Then, count how far that value is away from β and if you go the same distance in the other direction, you know the y-value

will be the same.

The best way to make this process clear is with some examples, so first

we will consider examples of parabolas that open up or down. Within

the examples, we will also review the concepts of domain, range, and

intercepts as we usually do.

π₯

π¦

(β, π)

The y values should match

for β + π and β β π ,

where π is the distance you chose to go

from β to plug in another value.

π π

539

Examples

For each of the following, graph the given relation. Give the vertex,

axis of symmetry, domain, range, as well as the x and y intercept(s).

1. π¦ = 3(π₯ β 1)2 β 2

First, you will notice that this example is already in standard

graphing form, so the square has already been completed. Recall

that standard graphing form is π = π(π β π)π + π.

We should be able to read off the vertex (β, π) as well as the axis

of symmetry π₯ = β pretty quickly from looking at the equation:

vertex: (1, β2)

axis of symmetry: π₯ = 1

We can start graphing by plotting the vertex:

π₯

π¦

Note again that we took the opposite

sign for what was inside of the

parentheses and the same sign for

what was outside.

540

Now decide what you want to plug in next. We will choose to

plug in 0 for π₯ (since that will also give us the y-intercept at the same time).

Plugging in 0 for π₯, we get:

π¦ = 3(0 β 1)2 β 2

= 3(β1)2 β 2

= 3 β 1 β 2

= 3 β 2

= 1

This means we have the point (0,1) on the graph as well:

Now to get the other side of the parabola, you can use symmetry.

Since you traveled a distance of 1 from the vertex to get to 0, you can go

a distance of 1 in the other direction to land at 2 and you should get the

same y-value. That means the point (2,1) will also be on the graph.

π₯

π¦

We can sketch half of the parabola

using the vertex and our new point,

although plugging in one more value

on this side will give you a more

accurate sketch. (e.g. if you plug in

β1, you will get 10)

541

Now, we can use this point to guide us when we graph the other half of

the parabola:

So our final graph (without the exaggerated points on it) looks like this:

We can see the domain and range pretty clearly from the graph, but we

also know the domain is all real numbers because this a polynomial

equation in the variable π₯ (we can plug in any real number for a polynomial variable and get out a real number). Following the graph

from bottom to top, we can see that the range is [β2, β) since the graph

π₯

π¦

π₯

π¦ There is a little shortcut that some teachers

like to show their students that allows them

to think of the value β²πβ² as a kind of βslopeβ,

but be careful not to confuse this with

graphing a line, because they are different

shapes!

With that in mind, the trick is to write β²πβ² as

a fraction and move up and over (in both

directions) from the vertex to get two

symmetric points on the parabola. In our

case, we would move up 3 and over 1 in

each direction to arrive at the points we

happened to get by plugging in a value and

using symmetry.

542

covers all of these π¦ values. We already have the y-intercept also since

we plugged in 0 earlier and found the point (0,1).

The only thing we donβt have yet are the x-intercept(s), and from the

graph we see that we should have two of them. To find them, we just

plug in 0 for π¦ and solve for π₯, as usual:

0 = 3(π₯ β 1)2 β 2

οο ο 2 = 3(π₯ β 1)2

οο ο 2

3 = (π₯ β 1)2

οο ο Β±β 2

3 = π₯ β 1ο

οο ο 1Β±β 2

3 = π₯

Now rationalize your answers:

οο ο π₯ = 1 Β± β6

3 ο ο

ο

or rounding them as decimals:

οο π₯ = 1.816